R.OUND AND 
OVAL CONES 



§£& 




JOHN FULLER, Sr. 







Class L 

Boot. Jl 



Copyright^ 



COPYRIGHT DEPOSIT; 



A New and Original Treatise 



ON 



The Geometrical Development 



OF 



Round and Oval Cones 



With Easy Examples of their Application 



FOR THE USE OF BEGINNERS AND 

PRACTICAL SHEET IRON AND 

TIN PLATE WORKERS 



By 
JOHN FULLER, SR., 

Author of The Art of Coppersmithing 



PUBLISHED BY 

DAVID WILLIAMS COMPANY, 
232 William Street, New York. 



T5 3U 



LIBRARY of CONGRESS 


Two Copies 


rtactsivcu 


mak a 


1905 


Oft*. *2, fqc¥ 

'eUiSS *s XAfc Not 
76/23 

COPY 8. 



Copyrighted. 1904, by John Fuller, Sr., Seneca. Kas. 



1 






DEDICATED 

TN loving memory to my wife, Ann, 
to whose faithful encouragement and 
patient example in difficulties I owe 
whatever success I have achieved during 
our -fifty years companionship in an 
eventful and busy life. 



Conical Vessels and Their Patterns. 

When the writer was a boy the problems involved in 
conical work were quite a puzzle to him, as they have been 
to many others who were in the unfortunate position of 
having no opportunity of gaining the necessary education. 
There were then no books or papers of a practical nature 
on these subjects within my reach. A boy under such 
conditions must either make friends of the men with 
whom his lot is cast or acquire proficiency without their 
help. I tried both expedients. Perhaps it will interest 
my boy readers to tell of my first lessons in conical 
work, and show their application in a few articles 
of every day use, because they furnish valuable hints 
for further search in the mysterious but interesting 
science of sheet metal working, or, more properly, prac- 
tical geometry. My first lessons in cones involved the 
making of common extinguishers and bedroom candle- 
sticks. These two articles were of much interest to 
me, an ambitious boy, at the threshold of the sheet metal 
trade. After I had been working two or three years, dili- 
gently investigating all the problems that came in my 
way, I found that the old workmen used five primary 
standard fashions in cones, after which they patterned 
their work, so that they could be easily understood when 
giving directions in the various kinds of conical work. 

These primary fashions were named and understood 
as follows: Extinguisher, muller, funnel, lantern-head 
and hood, and are illustrated in Figs. I, 2, 3, 4, 5. The 
envelope of the extinguisher, Fig. 1, is formed of one- 



6 The Geometrical Development 

sixth of a circle, or 60 degrees, and when turned into 
shape forms at the apex an angle of 20 degrees very 
nearly. The mull'er, Fig. 2, is formed of two-sixths of a 
circle, or 120 degrees, which, when turned into shape, 
forms at the apex an angle of 40, nearly. The funnel, 
Fig. 3, requires three-sixths, or 180 degrees, and makes, 
when turned into shape, an angle of 60 at the apex. 
The lantern-head, Fig. 4, takes four-sixths, or 240 de- 
grees, and forms an angle of about 80 when turned into 
shape ; while the hood, Fig. 5, takes five-sixths, or 300 
degrees, making an angle of no at the apex, approxi- 
mately. Within these five standard fashions once lay 
all the principal varieties of conical shapes used by the 
old workmen in sheet metal working requiring flaring 
sides. To explain: First, in A, Fig. 1, is represented the 
pattern of an extinguisher, once, used to put out the light 
of a candle. This first primary fashion gave the initial 
lesson in pattern cutting, and was an excellent boy's job. 
We were kept pretty busy in their manufacture. The 
writer made many a gross, and they afforded a good 
preparatory lesson in the art of turning by hand, as well 
as laying edges true and even, fit for soldering. To get 
the dimensions of this pattern we multiply the base by 
three ; this gives the radius of the circle, of which the 
pattern is a part. Thus: Suppose an extinguisher, A, 
Fig. 6, or cone for any similar article, is required an 
inch and a half in diameter at the base /; then 
1.5X3- 4-5, and one-sixth of a circle whose radius is 
4Y 2 - inches will make the extinguisher, without anything 
allowed, for lap, the tin coming together edge to edge. 
Next, in the same figure, B is an old-fashioned quart 



of Round and Oval Cones. 



Fig.l 



Fig.2 



Fig.3 





Ftg.4 





Fig.5 ! 



3 The Geometrical Development 

cup. This cup is made, it will be seen, extinguisher 
fashion, and the radius of the pattern is obtained in the 
same wav — namely, by multiplying the diameter of the 
bottom by three, which gives the radius of the circle of 
which the body of the cup is a part. Now, take for 
the length of the body one-sixth of the circumference of 
the circle, and for the depth of the cup one-third of the 
generating radius a b, thus: Suppose the bottom & c of 
the cup B is 4 inches in diameter ; then 4 X 3 — 12, the 
radius a b of the circle, one-sixth of whose circum- 
ference will be the length and form the outer edge of 
our pattern for the cup B, and one-third of 12, or 
4, will be the depth, also without edges for seam or 
wire. The pattern for the handle is shown in d, beside 
the cup. In C is shown a pretty milk can, cut in the same 
fashion as the quart cup, and the pattern, it will be seen, 
is obtained in the same manner, being one-third of the 
radius deep. The hasp e and catch for the cover are 
shown beside the can. 

These are three illustrations of this fashion with the 
base down, two of which are in thirds — that is, a third of 
the generating radius as the depth of the vessel. The 
next, Fig. 7, D, illustrates the same fashion in halves — 
that is, the article is made one-half the generating radius 
deep ; D represents a coffee pot with a strap handle and 
lip, and E is the same thing with a wood handle and 
spout. It can readily be seen that the same rule or 
fashion is used here, thus : If a coffee pot or any similar 
article measures 6 inches at the bottom, then 6 X 3 = 18, 
the radius of a circle of which one-sixth of the circum- 
ference is taken to form the body of the coffee pot, D, with 



of Round and Oval Cones, 





Fi?.6 




Fig. 



to The Geometrical Development 

one-half the generating radius taken for the depth. Many 
other examples could be given, but these are deemed 
enough to show the principle. In Figs. 8 and 9 are 
illustrated two common pails, one an open milk or water 
pail, the other a slop pail. Here the order is reversed 
in the vessel, and the small end is made to serve for the 
bottom, but the law is unchanged. The bodies of both 
are extinguisher fashion, as before, which can be seen 
at a glance, the depth of them in these cases being one- 
fourth of the radius of the circle of which their bodies 
are a part. This slop pail affords a good example or 
lesson for careful study, as it embraces three primary 
fashions in one vessel; thus, the booge (or breast) x is 
cut lantern-head ; the body y extinguisher, while the 
foot H is funnel fashion. Suppose now the pail, as 
shown in Fig. 8, is to be 10^ inches at the brim a b; then 
IO -5 X 3 = 31.5, the length of the radius a e, and if e a 
be divided into four equal parts we get in this instance 

the depth of the side, or — — - = 7.875 — th *t is, 7% 

without edges. Now, the slop pail body, Fig. 9, is the 
same as the milk pail, hence for the body at the section 
h g it is ioy 2 , and at the foot e d three-fourths of 10^, 
or y}i ; then e d, or 7%, is the diameter of the inside or 
small end of the foot H, and also the inner radius of 
the semicircular ring from which the foot H is formed 
funnel fashion. The foot may be any depth to suit the 
taste, although one-fourth the depth of the body, or 2 
inches in this case, is considered the right proportion. 
The booge (or breast) at h g is ioy 2 inches; then, 



of Round and Oval Cones. 




Fig.8 








1 / 






1 / 




\ 

€ 


1 / 

1 / ' 

/ ' 

/ 

/ 


/ 


s 


''\\ 






ja, 



Fig.9 



12 The Geometrical Development 

JO C Y 1 • 

— £ -=7.875, the radius kg, of which circle four- 
sixths is required to make the booge, which is also made 
one-fourth the depth of the pail wide when wired, and 
hollowed in the hollowing block to give it the required 
curve and make it ready for the pail. From the foregoing 
explanation, and accompanying familiar illustration, 
the reader may soon become acquainted with the general 
method of forming similar objects. 

MULLER FASHION. 

In Fig 10 are shown three examples in the next, or 
muller fashion, which is formed of 120 degrees, or 
two-sixths of a circle. This fashion in an ordinary 
tin shop is used for dish pans, A, and deep basins 
or pudding pans, B, or some other articles, as the 
grease kettle, C, all of which were once made by 
hand. When the fashion or standard shape has been 
determined, and we wish to make a dish pan or any 
similar vessel of a giver diameter, say, 12 inches, then 

TO "S/ "3 

we proceed thus: — j =18 inches, the radius of a 

circle, two-sixths of whose circumference will make 
the pan without edges. One-third the radius, or 6 inches, 
gives the proper depth. It should be noticed that many 
other vessels are made on exactly the same principle 
with the cone inverted. For example the grease kettle C, 
Fig. 10, is an exact pattern of a round bait kettle for fish- 

* The rule for obtaining the generating radius of any of these stand- 
ard fashions is : Multiply the diameter of the base of the cone to be 
formed by 3 and divide by the number of sixths of the circle used to 
form it. 



of Round and Oval Cones. 



13 




Fig.lO 




Fig. 12 



Fig.13 



14 



The Geometrical Development 



ing, and is the same fashion as the pans, but inverted, the 
base of the cone being turned down, while the depth is 
one-third of the generating radius, as before. The pan B 
may be made of any depth as may be required, its 
pattern being obtained in the same way, by multiply- 
ing the diameter of the brim by three and dividing the 
product thus obtained by two, when two-sixths of the 
circle of which the quotient is the radius will be the 
pattern required. 

FUNNEL FASHION. 

The funnel fashion, Fig. 3, is formed, as shown, of 
three-sixths, or one-half, of a circle, and is adapted es- 
pecially to the funnel, Fig. 3, and a few pans and similar 
flaring articles. Several applications of this fashion are 
here shown. Fig. 1 1 shows a spittoon, with the cone base 
down. Fig. 12 is a lamp filler, also with the base down. 
Fig. 13 is a pan with the base turned up, forming the 
brim, and Fig. 14 is a candlestick, the pan of which was 
once made in pieces. To obtain this pattern we take 
the diameter c' d! of the brim, Fig. 14, as the radius, and 
three-sixths of a circle whose radius is equal to the 
bottom or brim diameter will make the pattern required, 
the depth of the pan being added or subtracted from 
the radius as the case requires. Many other examples 
could be given to illustrate where this fashion can be and 
is used in the construction of many articles called for 
and made in a country shop, where most people expect 
to get anything made from sheet metal. 

LANTERN-HEAD FASHION. 

The lantern-head is formed of four-sixths of a circle, 
and was used for the tops of old-fashioned horn and 



of Round and Oval Cones. 



15 





'*-^s 



,^ to' 



1 6 The Geometrical Development 

mica lanterns, shown in Fig; 4; also oil bottles, Fig. 15; 
colanders, Fig. 17; wash dishes, Fig. 18. To obtain this 
pattern we proceed thus : If it is required to make a 
colander, Fig. 17, 12 inches at the brim a b, then the 

12 "X 'K 
rule is =9, and four-sixths of a circle whose 

radius, c d, is 9 inches, will make the colander, which 
should be one-half the radius, or 4.3/2 inches, deep — 
that is, without edges for seams or wire. Wash dishes, 
Fig. 18, sprinkler roses, Fig. 16, and all similar articles 
the same. Lantern-heads and oil bottle tops are ex- 
amples with the base of the cone down. The application 
of this rule to oil bottles or cans will next be shown. When 
the author made oil cans the tops were all made lantern- 
head fashion, and if the reader will examine the best 
specimens that come under his notice he will see those 
that please the eye the best are still constructed in ac- 
cordance with this standard fashion. A top made funnel 
fashion never looks well, because its form is not 
in harmony with the body. Let us make a gallon oil 
bottle, Fig. 15. We used to make gallon oil bottles from 
middle plates — that is, tin plates which measured 11 x 15 
inches after squaring up. Take, then, a middle plate 
and cut it in two, making the pieces 7^ x 11. Notch 
the two ends of each piece, as shown, and make the 
notch y% inch long the way of the seam and 3-16 the 
other, or one-half the groove, and fold them for groov- 
ing; then form them into cylindric shape by passing 
them through rollers, and groove down the seams; the 
two seams, it will be found, have taken up about £4 
inch, leaving us 21*4 inches in the circumference of the 



of Round and Oval Cones. 17 

21.2 K 

body. Then, — '—— = 6.764, or a little over 644 inches, 
J 3.1416 

the diameter of our oil can body. Now, burr both ends 

of the body a neat l /% inch and seam on the bottom. 

We are now ready for the top. Here we see the diameter 

of the base of our can top should be a little over 7 inches ; 

then, proceeding in accordance with the rule for this 

fashion, — = 5.25 ; that is to say, four-sixths of a 

4 

circle whose radius is 5*4 inches will make the top, as 

shown in Fig. 15. Now add enough on each side for 

seam parallel with the edges, as shown, and the pattern 

is complete, read)- for forming. After the top is formed 

and nicely rounded turn the edge at the base back 

twice the width of the burr on the body, and then turn 

up the edge for the seam. Next put in the neck for cork 

and rivet on the handle, snap on the top and pene it 

down for soldering. 

Let us now see if the can will hold a gallon. The 
number of cubic inches in an imperial gallon (English 
Standard) is 277.274. The body of the can is cylindrical 
and is 6.75 inches in diameter by 7.25 inches high. The 
volume of a cylinder is equal to the area of the base mul- 
tiplied by the hight, and the area is equal to the square 
of the radius of the base multiplied by 3.1416. Then 
3-375 X 3-375 X 3-1416 X 7-25 = 259-439 = the con- 
tents of the body part. Next the top is 6.75 inches in 
diameter and 3.7 inches high. The volume of a cone is 
equal to the area of the base multiplied by one-third of 
the altitude or hight. Then 3.375 X 3-375 X 3- I 4i6 X 



1 8 The Geometrical Development 

3.7 

— — 44.134 = the contents of the conical top and the 
o 

whole volume of the can is 

259439 X 44.134 = 303-573. 
which shows margin enough over 277.274 so that the can 
is of full capacity. The same formula used with an Amer- 
ican gallon (231 cubic inches) and a sheet 14 x 10 is a 
good example for practice. 

HOOD FASHION. 

The hood, or cap for stove pipe, is formed of five- 
sixths of a circle and is illustrated in Fig. 5. This fash- 
ion is used principally for making caps for stove pipe and 
flat covers, such as lard cans, spice boxes, bucket covers 
of different kinds, lamp crowns and many similar pieces 
used in the make-up of lamps and lanterns when they are 
made by hand. All these examples are instances of its 
use with the base down. Let it be required to make a 
hood for a stove pipe 11 inches in diameter, then, pro- 
ceeding by the rule in a similar way as before, we have 

11 X ^ 

— = 6.6, or 6 6-10 inches as the radius of a circle 

five-sixths of which will make the hood required, but 
without anything allowed for seam, which must be added 
parallel with the edge. The foregoing rules I have used 
for many years. They are very simple and practical, re- 
quiring but little thought or mathematical knowledge, and 
for the general run of work are sufficiently accurate. 
When the system is understood and fully grasped by the 
learner he will find it one of the most valuable systems 
for ready application in ordinary work. Sometimes, .how- 
ever, it happens that greater accuracy is necessary in 



of Round and Oval Cones. 19 





~ -''?5 



20 The Geometrical Development 

the premises, and we have an example which calls for 
clear reasoning. For instance, in Fig. 19 is an example 
which once presented itself to the writer, and perhaps it 
has also to the reader. It was required to make a large 
hood„ to hang over a smith's fire, ABC, Fig. 19, having 
an angle of 120 degrees at the apex, or a rise of 30 de- 
grees at or from the base. It was lined out on a board 
in a similar way to that for a longer taper, as in Fig. 19, 
and I was perplexed to find that the outlines a b c, 
c b d, d b a, or the three figures of the cone or hood re- 
quired, completed the circle. The question then was 
what part of the circle it was necessary to take out to 
make the cone required. In the emergency I could only 
cut and try, and study the matter out 'afterward. 

I waded through many books patiently searching for 
years for the required information, and finally found the 
key to it in an old mensuration book by John Bonny- 
castle, published in 1823, and here I present the result 
of my search, or its application to resolve cones of any 
given hight or rise from the base or angle at the apex 
accurately. To illustrate : Let it be required to make 
a cone, A B C, Fig. 20, 18 inches in diameter at the base, 
and any hight taken at random from the base to the 
apex, say, 8 inches perpendicularly, D B. Now we 
want to know how many degrees of a circle, whose radius 
is the slant hight, A B, of the cone A B C, it will take 
to make the cone. To do this it may be shown that 
where the radius of a circle is 1, half the circumference 

is 3. 141 59, and, therefore, J" = .01745, or the length 

of an arc of 1 degree. Hence, .01745 multiplied by the 



of Round and Oval Cones. 21 

number of degrees in the arc will give the length of arc. 
In the example before us A D = 9, and D B = 8; 
then n/ 9 2 + 8 2 = 12.0414, the radius A B ; then 
12.0414 X -01745 = .21012, or the length of an arc of 
1 degree of a circle whose radius is 12.0414. The diam- 
eter of the base circle A E C is 18; then 18 X 3-Hi6 = 

56.4488, and — =269.123, the number of degrees, 

or the length of that part of the circumference F A K 
C H G whose radius, A B, is 12.0414. To measure 
or cut off 269.123 degrees from the circle F A K C H G, 
which is 24.0828 inches in diameter, divide 269^ degrees 

260 12^ 

by 60 degrees, or — y ' ^ =4.485 — that is, four steps 

of 60 degrees and nearly one-half of another, which 

prove it to be half way between lantern-head and hood 

fashion. Now step off on the circumference with a 

pair of compasses four times the radius A B — that is, 

.0- 

F A, A K, K C, C H, and — of another space, as 

1000 ^ 

shown by F A K C H, and on to G, which will be the 

number of degrees required for the cone ABC, Fig. 20. 

One other example, Fig. 21. Let it be required to 

make a cone, ABC (or a frustum of a cone having the 

same slant or fashion), 18 inches in diameter at A B, and 

th e sla nt hight, AC, 15 inches; then D C = 

v/ (A C) 2 — (A D) 2 , or V 15- — g 2 = 12 2 ; therefore, 

D C, or the perpendicular hight, will be 12 inches. 

Now, the slant hight A C of the cone C A B — that 

is, the radius of the circle w x y 2 — is 15; then 

15 X -01745 = .26175, tne length of an arc of 1 degree 



22 The Geometrical Development 

of a circle whose radius is 15 inches, and the circum- 
ference of A B or base of the cone is, 18 X 3.1416 = 

56.5488, and \r — =216.0332, or the number of de- 
grees of a 30-inch circle necessary to make the cone 
A B C, which proves it to be between funnel fashion 
and lantern-head. Dividing 216.0332 degrees by 60 de- 

216.0332 r 

grees we get -?—$ = 3.0005, or three steps of the 

generating radius A C around the circle w x y z, and 
six-tenths of another step on to F, which measures off 
216.6 degrees, or a little more than half way between 
funnel fashion and lantern-head. 




To Construct Ovals. 

In the study of the properties of an oval or ellipse 
much time and labor are involved, which few men can 
spare or are willing to devote after a day's work. To 
know the ellipse one must inquire into the properties 
of an oblique section of a cylinder. In the follow- 
ing lessons treating of ovals and oval cones only those 
figures which can be drawn with a pair of compasses 
will be considered, the conical fashions coming in the 
same order as before in round cones and conical vessels. 
A few examples of ovals drawn with compasses, to in- 
troduce this curious but useful problem, will here be 
given. The last two examples shown are said to have 
been first proposed in 1774 by Rev. John Lawson, Rector 
of Swancombe, County Kent, England. All the other 
examples may be described with a pair of compasses, and 
are given as a prelude, the purpose being, as above stated, 
to treat only of those figures which may be drawn by 
this means. The first example, Fig. 22, is to construct an 
oval, the foundation of which is obtained from points 
on a square, thus : On the line A B construct the square 
A C D B, and divide the sides A B and C D into two 



Fig.22 



Fig.23 





24 The Geometrical Development 

equal parts at the points F and E; join F C and F D, 
also A E and E B, and describe the arcs A B and C D, 
with E and F as centers ; and with G A and H B as radii 
describe the arcs CIA and D J B, which complete the 
oval. 

The second example, Fig. 23, is to construct an oval 
from points on a square and a half. On A B erect the 
parallelogram A C D B, making A B equal 3 and A C 
equal 2. Divide the sides A B and C D into two equal 
parts at the points F and E ; join F C and F D, also E A 
and E B, and describe the arcs A B and C D, with E and 
F as centers, and with G A and H B as radii describe the 
arcs CIA and D J B, which complete the oval. 

The third example, Fig. 24, is to construct an oval 
from points on two squares. On A B erect the paral- 
lelogram A C D B, making A B equal 4 and A C equal 
2. Divide the sides A B and C D into two equal parts at 
the points F and E; join F C and F D, also E A and 
E B, and with F C and E B as radii describe the arcs 
A B and C D, and with G A and H B as radii describe 
the arcs CIA and D J B, which complete the oval. 

In the fourth example, Fig. 25, the foundation of the 
figure is obtained by or from two circles, the centers of 
which lay in a line passing through the transverse axis, 
A B, and whose conjugate axis passes through the in- 
tersection of the two circles at C and D. To describe this 
oval, draw the line A B and divide it into three equal 
parts, A E, E F, F B. Now, with the radius F B describe 
the circle BCD, cutting the circle A C D in C and D ; 
then with D G as radius from the points D and C describe 
the arcs G H and I J, and the oval is complete. 



of Round and Oval Cones. 



Fig.24 







26 The Geometrical Development 

In the fifth example, Fig. 26, the foundation of the 
figure is also obtained from two circles, the centers of 
which lay in the transverse axis of the figure, while the 
conjugate axis forms a tangent to each circle, at the 
center D of the transverse axis. To describe this oval 
draw the line A B and divide it into four equal parts, 
A C, C D, D E, E B. Now, with the diameter A D on 
C E as a common base, describe the two isosceles trian- 
gles C K E, C J E, extending the sides K C, K E and 
J C, J E. Then from C and E as centers with C D or 
E D as radius describe the arcs F A H and G B I, and 
from J and K with the radii J H and K F describe the 
arcs H G and F I, and the figure is complete. 

The sixth example, Fig. 2J, shows how to draw an 
oval when the length and width are given. Draw A B 
and D E, the length and width of the oval desired, at right 
angles to each other and cutting one another into two 
equal parts at the point of intersection, C. From A on 
A B mark off A G equal to D E, the width of the oval, 
and divide G B into three equal parts. With C as cen- 
ter and a radius equal to two of these parts, as G F, de- 
scribe arcs cutting A B in the points H and O. With H 
and O as centers and H O as radius describe arcs cutting 
each other in the points K and P. Join P H, H K, K O, 
O P ; in these lines extended the end curves will meet 
those of the "sides. With H and O as centers and H A as 
radius describe the end arcs, and with P and K as cen- 
ters and P D as radius describe the side arcs, and the 
oval is complete. 

If these six figures are studied and understood a good 
foundation is laid by which any oval may be constructed 



of Round and Oval Cones. 27 

that can be described by the aid of a pair of compasses. 
Further study of these figures will amply repay one for 
the time spent. 

TO FIND THE CIRCUMFERENCE OF AN OVAL, THE LENGTH 
AND WIDTH BEING GIVEN. 

Multiply the square root of half the sum of the 
squares of the two diameters by 3.1416, and the product 
will be the circumference, thus : Let the length be 8 and 
the width 6, then 



/g 2 _i_ 6* 

3.1416 V - =3. 1416 V/ 50=3. i4i6X7-07i= 22 -2i4. 

Rule 2. Multiply half the sum of the length and 
width by 3.1416 and the product will be the circum- 
ference near enough for most practical purposes. 

THE CIRCUMFERENCE BEING GIVEN, TO FIND THE LENGTH 
AND WIDTH OF OVALS. 

In Fig. 22 the ratio of the length to the width is as 
1 to .76394 — that is to say, if the length is 1 inch the 
width will be .76394 part of an inch. Let the circum- 
ference of Fig. 22 be 13.5 inches, then 13.5 X 2 -i- 5.5416, 
or (1 + .76394 X 3-i4i6) = 4.8722, the length, and 
4.8722 X 76394 = 3.722, the width — that is, an 
oval as in Fig. 22, whose circumference is 13.5 
inches, will be 4.8722 long and 3.722 wide. The proof is 

.11 13.S 1 4.8722 + 3.722 

thus shown: — ±£ ^-= 4.207, and-**— — — — =4.297. 

3.1416 • *' 2 

In Fig. 23 the ratio of the length to the width is as 

1 to .75 — that is to say, if the length is 1 inch the width 

will be .75 part of an inch. Let the circumference be 

13.5 inches, then 



28 The Geometrical Development 

" o = 4.91 1, or , °\, ^ = 4.91 1, 

5.4978 * y 1 + .75X3-1416 * y 

the length and 4.91 1 X 75 = 3.6832, the width — that is, 

an oval as in Fig. 23, whose circumference is 13.5 inches, 

will be 4.91 1 long and 3.6832 wide. The proof is thus 

1 J 3-5 * 4-9 11 + 3-6832 

shown : J D ^ = 4.297 and — — — — = 4.297. 

3.1416 y/ 2 *' 

In Fig. 24 the ratio of the length to the width is as 

1 to .7573 — that is, the length is 1 inch and the width 

-7573 P art °f an inch. Let the circumference be 13.5 

inches, then I3 ' 5 X 2 = 4.891, or, — p- 13 ' 5 ^ > = 

5.5207 * y ' ' 1 + .7573 X 3-1416 

4.891, the length, and 4.891 X -7573 = 3-7049, the width 

— that is, an oval as in Fig. 24, whose circumference is 

13.5 inches, will be 4.891 long and 3.7049 wide. The. 

proof is shown thus : 

13-5 j 4.891 + 3.7049 

3.1416 y/ 2 •* >i 

In Fig. 25 the ratio of the length to the width is as 

I to .7565. Let the circumference in this example be 

9 inches, then 9X2^ 5.182, or (1 + .7565 X 3.1416, 

the transverse axis added to the conjugate and multiplied 

by *) == 3.2619, the transverse axis, and 3.2619 X .7565 

— 2.4676, the conjugate — that is, an oval as in Fig. 25, 

whose circumference is 9 inches, will be 3.2619 long and 

2.4676 wide. The proof is shown as before, thus : 

3.2619 -f 2.4676 oz- j 9 o/r 

2_J ^-L — — 2.8647 and — ^— : = 2.8647. 

2 ^' 3.14 16- ^ 

In Fig. 26 the ratio of the length to the width is as 

f to .63238. Let the circumference in this example be 

13.5, then 13.5 X 2 — 27 and 27 -7- 5.128, or (1 + .6323X 



of Round and Oval Cones. 29 

3.1416) = 5.2652, the transverse axis, and 5.2652 X 
.6323 = 3.3291, the conjugate axis. The proof is 
S.26S2 4- 3.3291 . 13.5 

2 y/ 3.1416 ^ 7/ 

To Construct Oval Cones. 

Now to show the application of the five fashions as 
given for round cones, in the development of oval cones, 
some of which examples must at some time present them- 
selves to the active workman in his daily experience. 

EXTINGUISHER FASHION. 

Let it be required to make a cone extinguisher fash- 
ion, the base of which is an oval formed as in Fig. 28. 
First draw the oval as directed in Fig. 25, and let its 
transverse or longest diameter be 6 inches ; then the 
diameter of the circle A F will be 4 inches, and by the 

4X3 
rule give for extinguisher fashion — = 12, or the 

radius of a circle one-sixth of which would make a 
cone extinguisher fashion, whose diameter at the base 
will be 4 inches. But it will be seen that the oval is 
made up of six parts of circles, the radius of the middle 
or two side parts being twice the length of the radius 
of the four parts forming the two ends, and we want in 
constructing this elliptic cone six sectors arranged side 
by side whose radii are six times longer than those 
of which the oval in Fig. 28 is composed, but having 
their arcs equal in length to those of the oval, and equal 
when taken together to its circumference. Now con- 
struct the oval, Fig. 28, and with a radius three times 



3° 



The Geometrical Development 




N o' 



of Round and Oval Cones. 31 

the length of the diameter A F, from the point C as cen- 
ter, describe the arc L O K, making it 60 degrees, or one- 
sixth of the circle of which C K is the radius ; then from 
C through F and J draw the line C J to K, and from C 
through E I draw the line C I to L; now divide the arc 
L O K into six equal parts of 10 degrees each, L M N O 
PQK, and draw the line U C O, making U C = C O ; 
then draw N V and P T through C ; with C as a center, 
describe the arc T U V equal to N O P; from P with 
P C as radius describe the arc C Y, and from N with N C 
a9 radius describe the arc C Z, making C Y and C Z 
equal to N M and P Q ; then draw S P through Y, and 
W N through Z, and describe the arcs T S and V W 
with P and N as centers, and from the points Y and Z 
describe the arcs S R and W X, making them equal to 
L M and O K. Then the curve line R S T U 
V W X is equal to the circumference of the oval A H 
G B J I, and a pattern cut as indicated by R D X will 
make when formed a true elliptic cone extinguisher fash- 
ion, which may be used for slightly flaring oval vessels. 

MULLER FASHION. 

The muller fashion contains 120 degrees, or six sec- 
tors of circles which, when laid side by side, will equal 
the circumference of a given oval whose transverse 
diameter is A B, Fig. 29. Let A H G B J I, Fig. 29, be 
an oval 6 inches long, and let it be required to make a 
foot bath the shape of this oval, muller fashion ; then, by 
the rule given for muller fashion, three times A F divided 
by 2 equals the radius of a circle two-sixths of which 
would make a muller, Fig. 2, whose diameter at the brim 
would be 4 inches. But, as in the last example, we want six 



32 



The Geometrical Development 




OVAL BAIT PAIL 



of Round and Oval Cones. 33 

sectors whose radii are three times longer than those of 
which the oval is composed, but having their arcs equal to 
that of the oval, and which when all are taken together 
will be equal to its circumference. Now, then, with a 

radius equal to ^-or 6 (that is, 4 X 3 ~6), and 

from the point C as center describe the arc L O K, mak- 
ing it two-sixths of a circle of which six is the radius. 
Now divide the arc L O K into six equal parts of 20 
degrees each, as L M N O P Q K, and draw the line 
U C O, U C being equal to C O; then draw N V and 
P T through the point C; with C as a center describe 
the arc T U V equal to N O P ; now from P, with P C 
(that is, C T), describe the arc C Y, and from N, with 
N C (that is, C V), describe the arc C Z, making C Y 
and C Z equal to N M and P Q ; then draw S P through 

Y and describe the arc T S, from P, and draw W N 
through Z and describe the arc V W, from N, and with 

Y S and Z W as radii, from the points Y and Z describe 
the arcs S R and W X, making them equal to L M and 
Q K. Then the curve line RSTUVWXis equal to 
the circumference of the oval A H G B J I, and a pattern 
cut out as indicated by a R U X will make when formed a 
true elliptic or oval cone muller fashion, which may be 
used for large pans, foot baths and oval bait kettles and 
many other vessels. As shown in A and B, Fig. 29, A has 
the base turned up, and B has the base turned down. 

FUNNEL FASHION. 

Funnel fashion contains 180 degrees, or six sectors 
of circles which, when laid side by side, will equal the 
circumference of a given oval whose transverse diame- 



34 The Geometrical Development 

ter is A B, Fig. 30. Let A H G B J I, Fig. 30, be an 
oval 6 inches long, and let it be required to make a con- 
ical top funnel fashion to fit it; then, by the rule given 
for this fashion, three times A F divided by three equals 
the radius of a circle three-sixths of which would make 
a funnel, Fig. 3, whose diameter at the brim would be 
4 inches. But, similar to the two preceding examples, 
we want six sectors, whose radii are twice as long as 
those which compose the oval, but having their arcs equal 
to that of an oval, and which, when all taken together, 
will be equal to its circumference. Then, with a radius 

equal to — = 4 (that is, 4X 3 = 4), from the 

point C as center describe the semicircle L O K; 
now divide the semicircle into six equal parts of 30 
degrees each, L M N O P Q K, and draw the line 
O C U, O C being equal to C U; then draw I V and 
J T through C, and describe the arc T U V equal to L O K 
from C as a center ; now from J with J C as radius (that 
is, C T) describe the arc C Y, and from I with I C as 
radius (that is, C V) describe the arc C Z, making 
them equal to I M and J Q; then draw S J through Y, 
and with J T as radius describe the arc T S, and draw 
W I through Z, and with I V as radius describe the arc 

V W, and with Y S and Z W as radii, from the points 

Y and Z describe the arcs S R and W X equal to 
M L and Q K ; then the curve line RSTUVWXis 
equal to the circumference of the oval A H G B J I, as 
before, and a pattern cut as indicated by R U X will, 
when formed, make a true elliptic cone funnel fashion, 
which may be used for milk pans, dish pans and many 
other vessels for which it may be suited. 



of Round and Oval Cones. 35 




36 The Geometrical Development 

LANTERN-HEAD FASHION. 

The lantern-head fashion is made up of 240 degrees 
and contains six sectors of circles which, taken to- 
gether, will equal the circumference of a given oval 
whose transverse diameter is A B, Fig. 31. Construct 
the oval A H G B J I, making the length 6 inches, as in 
the preceding example, and divide the transverse axis 
A B into four equal parts, A E, E b, b F, F B, and draw 
U C D at right angles to A B ; then from the point C, with 
any radius, C O, describe the part of a circle L O K, 
making the length of the arc O L two-sixths, and O K 
also two-sixths, or the whole part, L O K, 240 degrees. 
Now divide LOK into six equal parts, L M N O P 
Q K, and from N, through E and C, draw N E V, and 
from P, through F and C, draw P F T. By the rule for 

, , AZX3 4X3 

lantern-head = 3, or = 3, the radius 

4 4 

of a circle four-sixths of which would make a lantern- 
head, Fig. 4, whose diameter at the base would be 4 
inches; then, with C T or 3 as radius, from the point C 
as center describe the arc T U V. Now on the line 
TCP from the point C lay off the distance C d equal 
to C T, and on the line V C N lay off the distance C h 
equal to C V, and from d and h, with the radii d C and 
h C, describe the arcs C Y and C Z, equal to T U and 
U V, and through the points Y and Z draw d Y S and 
h Z W ; then from d as center with d T as radius describe 
the arc T S, and from h as center and h V as radius 
describe the arc V W ; now from the points Y and Z, with 
the radii Y S and Z W (that is, C T), describe the arcs 



of Round and Oval Cones. 



3/ 



S R and W X, equal to T U and U V, and curved line 
RSTUVWXis equal to the circumference of the 
oval A H G B J I, and a pattern cut as indicated by 
b R U X will, when formed, make a true elliptic cone 
lantern-head fashion, which may be used for shallow 
pans, low bottle tops or for pieced dish covers. 




38 The Geometrical Development 

HOOD FASHION. 

The hood is the lowest or flattest when formed of all 
the five primary fashions, and is not much used, although 
it would be quite useful for wash boilers if the boiler 
was made elliptical. It is made up of 300 degrees and 
contains six sectors of circles which, when taken to- 
gether, will equal the circumference of a given oval 
whose transverse diameter is A B, Fig. 32. Construct 
the oval A H G B J I, shown by the dotted circumference, 
and make it 6 inches long, as in the preceding example. 
Then from the point C, with any radius, C O, describe 
the circle L O K, and having marked off five-sixths of 
its circumference, as L O K, divide it into six equal 
parts, as shown, by L M N O P Q K. From N (which 
marks off one-sixth of these five-sixths from the point 
O), through C, draw N C V. By the rule for hood 

A h X S 

fashion = 2.4, or the radius of a circle five- 

5 
sixths of which will make a cone hood fashion 4 inches 

4 X 3 
in diameter at the base, or — = 2.4, or the radius 

5 . 

C T or C V. Then with C T as radius describe the arc 

T U V. With C T from C on the line TCP mark off 

the distance C F, and with C V from C on the line V C N 

mark the distance C E, and with E as center and E C 

as radius describe the arc C Z, equal to U V, and with 

F as center with F C as radius describe the arc C Y, equal 

to T U, and from E, through Z, draw E Z W, and with 

E V as radius the arc V W, and from F through Y draw 

FYS, and from the point F as center and F T as radius 



of Round and Oval Cones. 



39 



describe the arc T S. From Y, with Y S as radius, 
describe the arc S R and make it equal to T U, and from 
Z with Z W describe the arc W X and make it equal to 
V U. Now join Y R and Z X, continuing them to their 
intersection a, and the curve line R S T U V W X will 
equal the circumference of the oval A H G B J I, and 
a metal pattern cut as indicated by a R U X will make 
a true elliptic cone hood fashion, which, when formed, 
will stand on the oval shown by the dotted lines in Fig-. 32. 

Note. — In the formation of elliptic cones, it should be observed 
that at the points C and Z where the foci vanish toward the apex a 
the pattern should be formed sharp to preserve the truth of the base oJI 
the cone. 




4© The Geometrical Development 

OVAL CONES (CONTINUED). 

We will now consider the five fashions of cones to 
suit the oval illustrated by Fig. 26. The foundation of 
this figure is two circles, the centers of which lay in the 
longest diameter, the shortest diameter forming a tangent 
to these circles in the center of the longest diameter of 
the oval, Fig. 26. 

EXTINGUISHER FASHION. 

Let it be required to make a cone extinguisher fash? 
ion, the base of which is an oval formed as in Fig. 33. 
First draw the oval A H G B J I as directed in Fig. 
26, making the longest diameter, A B, 6 inches; then 
the diameter of the circle A b will be 3 inches, and the 

rule for extinguisher fashion is — = 9, radius of . 

a circle one-sixth of whose circumference would make 
a cone of this kind of taper or extinguisher fashJon — that 

is, A b = 3 and 3 * 3 = g . t j ien q tj equals 9 inches. Now 

draw U C O, making C O equal to C U, and with C O 
describe the arc LOK, making it equal to one-sixth of the 
circle of which it is a part, and then divided into six 
equal parts, as L M N O P Q K. From P through 
C, draw T C P and extend it on to d, making T d three 
times the length of C U, or 2J inches long, and from V, 
through C, draw V C N and extend it on to h, making 
V h three times the length of C U, or 27 inches long. 
Then with C T describe the arc T U V. Through O 
draw the lines JOS and with h O W and with d T de- 
scribe the arc T S and with h V the arc V W. From the 
point h with the radius h C describe the arc C Z, and 



of Round and Oval Cones, 



41 




4 2 The Geometrical Development 

from the point d with the radius d C describe the arc C Y. 
From the points Y and Z with the radii Y S and Z W 
(that is, C U, or 9 inches) describe the arcs S R and 
W X, making them equal to T U and U V, and draw R a 
through Y and X a through Z. Then the curve line R S 
T U V W X is equal to the circumference of the oval 
A H G B J I, and a pattern cut as shown by a R U X will, 
when formed, make a true oval cone extinguisher fashion. 
To further explain to the learner the foregoing prin- 
ciples : If a straight line be drawn from R to X, then R a X 
will be an equilateral triangle, and the angle at a will 
be 60 degrees ; but the curved line is made up of sectors 
of two distinct circles, the radius of the larger one being 
three times the length of the other. The reason for this 
is because the radius of the two side arcs of the oval 
A B is three times longer than the radius of the arcs of 
the two ends, and, therefore, this is necessary to preserve 
the symmetry and proportion of the conic curve. Again, it 
will be seen by a little study that the angle contained in 
each sector of the pattern here developed is six times 
less than that of the oval from which the pattern has 
been evolved, while, on the other hand, the radii of the 
sectors are six times longer than those which compose 
the oval A B. In the next example, Fig. 34, it will be 
shown that in the development of the pattern the two 
radii forming it have been reduced by the rule one- 
half, while the angles of the sectors have been doubled — 
that is, when compared with the last example. This law 
must follow in proportion as the hight of the cone re- 
cedes, and having vanished has become a plane and the 
sectors coincide with those of the oval. 



of Round and Oval Cones 43 

MULLER FASHION. 

Let it be required to make a cone muller fashion to 

fit any size oval, as in Fig. 34. Construct the oval A H G 

B J I and let the length A B equal 6 inches. By the rule 

A. b X 3 

' = 4.5, the radius of the circle two-sixths of 

which will make a cone nauller fashion — that is, — — 

2 

4.5 ; then C U equals 4.5, and by the same rule 
"CJ -fCJX3 = I3 5j Qr T p Draw xj C o at right 

angles with A B, and with C U as radius from the point 
C describe the arc T U V. Now, with C P as radius from 
the point C describe the arc L O K, making it 120 de- 
grees, or two-sixths of the circle of which C P is the 
radius, and divide it into six equal parts, L M N O P Q K, 
three from O to L and three, from O to K ; then from P 
through C draw P C T and from N through C draw 
N C V. With P T as radius from the point T describe 
the arc T S, and with N V as radius from the point N 
describe the arc V W,and from the same centers the arcs C 
Y and C Z, making them equal to P O and N O ; then from 
P through Y draw P Y S, and from N through Z draw 
N Z W. From the points Y and Z with the radii Y S 
and Z W equal to C U describe the arcs S R and W X, 
and draw R Y and X Z and continue them until they 
meet in a; then the curve line RSTUVWXis equal 
to the circumference of the oval A H G B J I, and a 

* The diameter being twice the radius. 



44 



The Geometrical Development 



pattern cut as represented by R U X a will, when 
formed, make a true oval cone muller fashion. 



Fig.34 




- K 



LofC. 



of Round and Oval Cones. 45 

FUNNEL FASHION. 

About 50 years ago we made what were called break- 
fast bottles — that is, oval tin bottles to carry tea or 
coffee in for breakfast. The shape of the bottle was an 
oval, as shown in Fig. 26. The body was mounted with 
an oval conical head or top, the pattern of which was 
supplied to us, as was also that of the neck and handle, all 
of which were kept together in the usual way on a 
ring, and we had no trouble. But after leaving the old 
shop I was called on to make some bottles where there 
were no patterns, and here it happened, as it has often 
happened with others, my trouble began. I, however, 
persevered through the job as best I could, and then spent 
many weary hours trying to find some correct way which 
could be relied on and easily retained in memory, ready 
for use at any time ; at length I found what seemed the 
perfect plan. At this time I knew nothing of geometry ; 
I had never seen " Euclid's Elements " even, or any other 
book of the kind, but was eagerly plodding along trying 
to add to the practical instruction my father was able to 
give me ; and while I have somewhat extended the ideas 
then caught as they flitted by, I have never improved 
them in the main, because the first deductions were nearly 
geometrically true. 

To illustrate my first discovery : Let it be required to 
make an oval bottle to hold a quart nominally; then, 
one-half a single plate (sheet 10 x 14 inches) cut length- 
wise, with the seam deducted, or 13.5 long, will when 
formed into a cylinder hold a quart, and will measure 
4.2971 in diameter and 5 inches deep. But the bottle 



46 The Geometrical Development 

is to be an oval form, as shown in Fig. 26, and the ratio 
of the length to the width of this oval or transverse axis 
is to the conjugate as 1 is to .63238, or if the length is 
1 the width will be .63238. Now we must know what 
the length and width of an oval, as in Fig. 26, will be 
whose circumference is 13.5 inches. To do this we proceed 
as follows: Multiply the circumference by 2, and divide 
by 5.128 + (which is 1 + .63238 X 3-i4 I 6, or the trans- 
verse axis added to the conjugate and multiplied by 
3.1416), thus : 13.5 X 2 = 27, and 27 -7- 5.128 = 5.2652, 
the transverse-axis, and 5.2652 X .63238 = 3.3291, the 
conjugate axis. The proof is shown by addingthe length and 
width together and dividing by 2, which gives the circular 

diameter, thus: — — "1" 4-4 9 = 4,2971. Then by 

turning the strip into a circle we have — iLi ^-= 4.2971, 

which coincide near enough for all practical purposes. 
We now have the length and width of the oval, whose 
circumference is 13.5 — namely, 5.2652, the length, and 
3.3291, the width. Now lay off the length of the oval 
A B, Fig. 35, making it 5.2652 long, and draw C D at 
right angles to it at the point E. Now divide A E in F 
and E B in G, and with G B as radius and G and F as cen- 
ters describe the arcs H B I and K A J and let them 
each contain two-sixths, or 120 degrees. From I 
through G draw I G M, and from J through F draw 
J F M, and with M I as radius from the point M de- 
scribe the arc J I. From K through F draw K F L, and 
from H through G draw HGL, and with L K as radius 
from the point L describe the arc K H, and the oval 



of Round and Oval Cones. 



47 



Fig.35 




48 The Geometrical Development 

A J I B H K is complete and is the pattern of the bottom 
of the bottle without edges for seam, and from the lines 
of this bottom we develop the top or oval cone, proceed- 
ing as follows : Extend the line M F J both ways to P and 
O, and M G I to Q and N, and with the radius M P (that 
is, E B, which is twice the length of G B) describe the arc 
Q P ; now with the distance E B from the point G lay off 
the distance G N, and with the distance A.E from the 
point F lay off the distance F O, then from O through G 
draw the line O R, and from N through F draw the line N 
S, and with N S as radius (which is twice the length 
of L K) from the point N describe the arc Q S ; also 
with O R as radius (which is twice the length of L H) 
from the point O describe the arc P R ; now with O M 
describe the arc M U, and with N M describe the arc 
M T, then on the line N S with the distance T S (that 
is, A E) as radius from the point T describe the arc 
S W, and on the line O R with the distance U R (that 
is, E B) as radius from the point U describe the arc R V. 
Then through T and U draw the line W T U V, and the 
curve line W S Q P R V is equal to the circumference 
of the oval A K H B I J. The semicircle on X Y Z 
measures the hole for the neck of the bottle. The pattern 
here described as funnel fashion is true in every par- 
ticular. If the curve was a complete semicircle instead 
of an irregular curve and drawn from the center Y it 
would be an exact pattern of a common funnel. All the 
radii by which it has been constructed are double the 
length of those which form the oval from which it has 
been developed. 



of Round and Oval Cones. 49 

LANTERN-HEAD FASHION. 

Lantern-head fashion, as previously shown, is made 
up of 240 degrees. To make this pattern for any oval 
shaped as in Fig. 36 construct the oval A H G B J I 
and let A B equal 6 inches and draw U C D at right 

A d V 7. 

angles to A B. Then equals the radius of a 

circle four-sixths of which would make a cone lantern- 
head fashion— that is, 3 3 = 2.25, or C U. With any 

4 
radius, C O, describe the part of a circle, L O K, and 
make the arcs O L and O K together equal to four-sixths 
of the circle of which C O is the radius. Now divide 
L O K into six equal parts, L M N O P Q K. From 
P through C draw P C T, and from N through C draw 
N C V. Then from the point C with C U as radius 
(that is, 2.25) describe the arc T U V. On T C P from 
C lay off the distance C F twice the length of C T, and 
on V C N from C lay off the distance C E twice the 
length of C V. From F through Y draw FYS, and from 
F with F T as radius describe the arc T S. From E 
through Z draw E Z W, and from E with E V as radius 
describe the arc V W. With E C as radius from the 
point E describe the arc C Z, and with F C as radius from 
the point F describe the arc C Y. Now from the point 
Y with Y S as radius (that is, C T) describe the arc S R, 
equal to T U, and from the point Z with Z W (that is, 
C V) describe the arc W X, equal to V U, and draw 
RYaandXZ a. Then the curve line R S T U V W X 
is equal to the circumference of the oval A H G B J I, 
and a pattern cut as shown by this curve line will, when 
formed, make a true oval cone lantern-head fashion. 



5o The Geometrical Development 




of Round and Oval Cones. 51 

HOOD FASHION. 

Hood fashion, as before shown, is composed of six 
sectors which, when laid together, as in Fig. 37, will 
equal 300 degrees at the center a. To make this pattern 
fit any oval shaped as in Fig. 37 construct the oval 
A H G B J I and let the length A B equal 6 inches, and 
draw U C D at right angles to A B. Then A b equals 

3, and by the rule 3 3 = 1.8, or C U. With any radius, 

C K, describe the part of a circle, L O K, and make it 
equal to five-sixths, or 300 degrees. Now divide it into 
six equal parts, L M N O P Q K. From P through C 
draw P C T, and from N through C draw N C V ; then 
from the point C with C U as radius (that is, 1.8) de- 
scribe the arc T U V. On TCP from C lay off the dis- 
tance C h twice the length of C T, and from the point h 
with h T as radius (that is, 1.8 X 3, or 5.4) describe the 
arc T S, making it equal to one-sixth of a circle of which 
D H is the radius, and draw h Y S ; then with h C as 
radius describe the arc C Y. On N C V from C lay off the 
distance C d twice the length of C V (that is, 5.4), 
and from the point d with d V as radius, describe the 
arc V W, making it equal to one-sixth of a circle of which 
D H is the radius, and draw d Z W; then with d C as 
radius describe the arc C Z. Now from the point Y with 
Y S (that is, C T) describe the arc S R, making it equal 
to T U, and draw R a through Y, and from the point Z 
with Z W (that is, C V) describe W X, making it equal 
to V U, and draw X a through Z. Then the curve line 
RSTUVWXis equal to the circumference of the 



52 



The Geometrical Development 




of Round and Oval Cones. 53 

oval A H G B J I, and a pattern cut as shown by the 
curve line will, when formed, make a true oval cone hood 
fashion. 

In concluding the lessons on oval cones, a few words 
are necessary, as before noted, to show that in order to 
preserve for any special purpose the truth of the base 
when formed the pattern should be bent sharp along the 
lines a C and A Z, because the centers or the foci of the 
oval vanish at these points and meet in the center a, 
which will be readily seen by a few trials in practice. 



MAR 3 1905 



